Let $g$ be a vector-valued function defined by $g(t)=(t^2-1,2\sqrt{t})$. Find $g'(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $2t-\sqrt{t}$ (Choice B) B $\left(2t,\dfrac1{\sqrt t}\right)$ (Choice C) C $\left(2t,-\dfrac2{\sqrt t}\right)$ (Choice D) D $\left(\dfrac{t^3}{3}-t,3\sqrt {t^3}\right)$
Answer: $g$ is a vector-valued function. This means it takes one number as an input $(t)$, but it outputs two numbers as a two-dimensional vector. Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as $u(t)=(v(t),w(t))$, then its derivative is the vector-valued function $u'(t)=(v'(t),w'(t))$. In other words, the derivative is found by differentiating each of the expressions in the function's output vector. Recall that $g(t)=(t^2-1,2\sqrt{t})$. Let's differentiate the first expression: $\dfrac{d}{dt}(t^2-1)=2t$ Let's differentiate the second expression: $\dfrac{d}{dt}(2\sqrt{t})=\dfrac1{\sqrt t}$ Now let's put everything together: $\begin{aligned} g'(t)&=\left(\dfrac{d}{dt}(t^2-1),\dfrac{d}{dt}(2\sqrt{t})\right) \\\\ &=\left(2t,\dfrac1{\sqrt t}\right) \end{aligned}$ In conclusion, $g'(t)=\left(2t,\dfrac1{\sqrt t}\right)$.